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Greeley [361]
3 years ago
15

A group of students wants to determine the internal resistance of a battery. They connect the battery to a variable resistor. Th

e students measure the potenial differnce across the battery as a function of the current throught the battery as they vary the resistance. Which of the following analyses of the data could be used to determine the internal resistance of the battery?
A. Divide the potential difference across the battery by the current through it for each data point. The average of these calculations gives the internal resistance of the battery.
B. Graph the potential difference across the battery as a function of the current through it. Extrapolate to find the y- intercept and divide this by the average of the current measurements to find the internal resistance of the battery.
C. Find the best-fit straight line for a graph of potential difference across the battery as a function of the current through it. The absolute value of the slope represents the internal resistance of the battery.
D. This data cannot be analyzed to give the internal resistance of the battery, because the potential difference across the battery does not depend on the current.
Physics
1 answer:
andre [41]3 years ago
5 0

Answer:

The answer is "Option C"

Explanation:

Ohm's Rule we remember V = I\times R

It looks like the sample is drawn y=mx while m is curve slope

The slope of the graph is its resistance if they draw a graph v vs I.

However, the internal battery presence resistance can be modified as the straight line equation

y = mx -c

where c is y-intercept:

E = I(R+r) \\\\ E = IR+ Ir\\\\E = V + Ir\\\\V = E- Ir\\\\V = -Ir+E

R is really the internal battery resistance to draw graph V vs I, the slope is internal resistance to the negative slope cure.

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valentina_108 [34]

Answer:

The sled needed a distance of 92.22 m and a time of 1.40 s to stop.

Explanation:

The relationship between velocities and time is described by this equation: v_f=v_0+a*t, where v_f is the final velocity, v_0 is the initial velocity, a the acceleration, and t is the time during such acceleration is applied.

Solving the equation for the time, and applying to the case: t=\frac{v_f-v_0}{a}=\frac{0\frac{m}{s}-282\frac{m}{s}  }{-201\frac{m}{s^2} }=1.40s, where v_f=0\frac{m}{s} because the sled is totally stopped, v_0=282\frac{m}{s} is the velocity of the sled before braking and, a=-201\frac{m}{s^2} is negative because the deceleration applied by the brakes.

In the other hand, the equation that describes the distance in term of velocities and acceleration:x_f-x_0=v_0*t+\frac{1}{2}*a*t^2, where x_f-x_0 is the distance traveled, v_0 is the initial velocity, t the time of the process and, a is the acceleration of the process.

Then for this case the relationship becomes: x_f-x_0=282\frac{m}{s} *1.40s+\frac{1}{2}(-201\frac{m}{s})*(1.40s)^2=94.22m.

<u>Note that the acceleration is negative because is a braking process.</u>

4 0
4 years ago
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A bowling ball is launched from the top of a building at an angle of 35° above the horizontal with an initial speed of 15 m/s. T
Mamont248 [21]

Let y_0 be the height of the building and thus the initial height of the ball. The ball's altitude at time t is given by

y=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ\,t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity.

The ball reaches the ground when y=0 after t=2.9\,\mathrm s. Solve for y_0:

0=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ(2.9\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(2.9\,\mathrm s)^2

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3 0
3 years ago
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Answer:

Work = 5941 J

Explanation:

As we know that work done is given by the equation

W = F. d

here we know that

F  = (200 N)\hat i - (149 N) \hat j

also we have

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W = (200 N\hat i - 149 N \hat j).(23 m\hat i - 9m \hat j)

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Answer:

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Explanation:

The density of an object is given as the ratio between the mass of that object and the volume occupied by that object.

Density = Mass/Volume

Now, it is given that the density of Earth has become:

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Mass = Mass of Earth (Constant) = 5.97 x 10²⁴ kg

Volume = 4/3πR³ (Volume of Sphere)

R = Radius of Earth = ?

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4/3πR³ = (5.97 x 10²⁴ kg)/(1 x 10⁹ kg/m³)

R³ = (3/4)(5.97 x 10¹⁵ m³)/π

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